Question

Please solve it!!
The function g is defined by $\sf g(x)=q+\dfrac{P}{x}$, $\sf g^2(x)=g^{-1}(x)$. Prove that P+q²=0.

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Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{g(x) = q + \dfrac{p}{x} } \\ &\sf{ {g}^{2}(x) = {g}^{ - 1}(x)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: prove - \begin{cases} &\sf{{q}^{2} + p = 0}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\large \underline{\tt \: \red{ According \: to \: statement }}$

$\tt \longmapsto\:g(x) = q + \dfrac{p}{x}$

Now,

• Let first evaluate inverse of g(x).

$\tt \longmapsto\:let \: g(x) = y = q + \dfrac{p}{x}$

$\tt \longmapsto\:y - q = \dfrac{p}{x}$

$\rm :\implies\:x = \dfrac{p}{y - q}$

$\rm :\implies\:\:\boxed{ \green{ \bf \: {g}^{ - 1}(x) = \dfrac{p}{x - q}}} - - (1)$

Now,

• Let evaluate,

$\tt \longmapsto\: {g}^{2} (x)$

$\tt \longmapsto\: = g(g(x))$

$\tt \longmapsto\: = g\bigg( q + \dfrac{p}{x} \bigg)$

$\tt \longmapsto\: = q + \dfrac{p}{q + \dfrac{p}{x} }$

$\tt \longmapsto\: = q + \dfrac{px}{xq + p}$

$\tt \longmapsto\: = \dfrac{pq + x{q}^{2} + px }{qx + p}$

$\rm :\implies\:\:\boxed{ \green{ \bf \: {g}^{2}(x) = \dfrac{pq + x{q}^{2} + px }{qx + p} }}$

Now,

$\large \underline{\tt \: \red{ According \: to \: statement }}$

$\tt \longmapsto\: \blue{ {g}^{2}(x) = {g}^{ - 1}(x) }$

$\tt \longmapsto\:\dfrac{pq + x{q}^{2} + px }{qx + p} = \dfrac{p}{x - q}$

$\tt \longmapsto\:xpq + {x}^{2} {q}^{2} + {px}^{2} - {pq}^{2} - {xq}^{3} - xpq = xpq + {p}^{2}$

$\tt \longmapsto\: {x}^{2} {q}^{2} + {px}^{2} - {pq}^{2} - {xq}^{3} - xpq - {p}^{2} = 0$

$\tt \longmapsto\: ({x}^{2} {q}^{2} + {px}^{2}) - ({pq}^{2} + {p}^{2}) - ({xq}^{3} + xpq) = 0$

$\tt \longmapsto\: {x}^{2}( {q}^{2} + p) -xq( {q}^{2} + p) - p( {q}^{2} + p) = 0$

$\tt \longmapsto\:( {q}^{2} + p)( {x}^{2} - xq - p) = 0$

$\rm :\implies\:\:\boxed{ \green{ \bf \: {q}^{2} + p = 0}}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$